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# The Inverse Tangent Function - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I want to graph a transformation of the inverse of the tangent function. Let’s graph y equals negative inverse tangent x plus pi over 2. Now whenever you’re doing a transformation graph, you want to have some key points that you can use in your transformation and if you’re not 100% sure of what points to use, remember that the inverse tangent function is the inverse of the restricted tangent function.

Let’s take a look at some points here. These three points of the tangent function are the ones we would use if we were graphing a transformation of tangent. They are negative pi over 4, -1, here 0,0 here and here we have pi over 4,1. For inverse tangent I would use the interchange of these. You switch the x and the y coordinates, so I’d get negative 1, negative pi over 4, again 0, 0 and 1, pi over 4.

The other thing I have to keep track of because I’m basically transforming this blue graph are its two horizontal asymptotes. It’s the asymptote at y is equals pi over 2 and y equals negative pi over 2. I can apply the transformations to those, those asymptotes as well. So we’ve got negative inverse tangent x plus pi over 2.

Now what kinds of transformations do these numbers represent? The -1 represents a reflection across the x axis. And to get that I take these y values and I multiply them by -1 and the plus pi over 2 represents a shift upwards by pi over 2 units and I can get that by adding pi over 2 to the result. I take each of these numbers, multiply by -1 and then add pi over 2. Let’s do that. So negative pi over 4 times negative one is pi over 4 plus pi over 2 is 3 pi over 4. Times -1 plus pi over 2 is pi over 2. Times -1 plus pi over 2 is pi over 4. And when you look at this function, there’s no actual horizontal transformation going on, so the x values will be unaffected.

So I just translate these over, -1, zero and 1. This gives me three points that I can plot for my new transformed function. Negative one, 3/4pi, zero pi over 2 and 1 pi over 4. Let’s plot those.

Negative 1,3 pi over 4 would go about here, 0,pi over 2 and 1 pi over 4. Now what happens to the asymptotes? Well I can apply all the transformations I applied to these y values to the asymptotes as well because they’re y values as well. So I multiply these by -1 which basically reverses them and then I add pi over 2. So this first one becomes times -1 gives me negative pi over 2 plus pi over 2 gives me y equals zero. Times -1 gives me pi over 2 plus pi over 2 gives me y equals pi. These are my two new asymptotes. Let me plot those. Y equals pi and y equals 0. And now I’m ready to finish the graph. Don’t forget to show your asymptotic behavior.

Now this is the graph of y equals negative inverse tangent x plus pi over 2 and the reason this graph is so interesting is that it's actually identical to inverse cotangent. Which suggests the identity; inverse cotangent x equals negative inverse tangent x plus pi over 2. This identity is actually related to the co-function identity for the tangent and cotangent.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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